Chapter C, Problem 36E

To determine

To calculate:

The value of $x$ to satisfy the following inequality equation:

  $\sin x>\cos x$

Answer to Problem 36E

The values of $x$ for the interval $\left[0,2\pi \right]$ to satisfy eq. $\sin x>\cos x$ are $\left[\frac{\pi }{4}\le x\le \frac{5\pi }{4}\right]$

Explanation of Solution

Given information:

  $\begin{array}{l}\sin x>\cos x\\ \left[0,2\pi \right]\end{array}$

Calculation:

Consider;

The unit circle with co-ordinates $\left(x,y\right)=\left(\cos x,\sin x\right)$ and radius $1$ to find out the values of $x$ for the interval $\left[0,2\pi \right]$ that cause $\sin x>\cos x$ .

  

Fig. Unit circle

From the figure;

  $\sin x>\cos x$ , in the second quadrant $x$ is negative and $y$ is positive. So, in second quadrant the given inequality equation is satisfied.

In the fourth quadrant $\sin x>\cos x$ , no value of $x$ satisfy the inequality equation because sine is negative and cosine is positive here.

In the first quadrant, $\left[\frac{\pi }{4}\le x\le \frac{\pi }{2}\right]$ and in third quadrant $\left[\pi \le x\le \frac{5\pi }{4}\right]$

Therefore, the values of $x$ for the interval $\left[0,2\pi \right]$ to satisfy eq. $\sin x>\cos x$ are $\left[\frac{\pi }{4}\le x\le \frac{5\pi }{4}\right]$